We immediately get two other points C and D symetric of A and B from axis.

Considering a 5

AB and CD intersect in O.

AC and parallel from B to axis (line B to P∞) intersect in U.

OU and parallel from D to axis (DP∞) intersect in V.

AV is the tangent in A to parabola.

Angle reflexion of the parallel to axis on this tangent intersects axis in the focus F.

The directrix is the perpendicular to axis from point H, symetric of F from the tangent.

We then construct immediately a 4

Considering then the 5

BC and AD intersect in U.

AB and parallel to axis from D (DP∞) intersect in V

UV and parallel to axis from C (CP∞) intersect in W

AW is the tangent in A.

Do the same for constructing the tangent in B, reversing the role of A and B.

Reflexions on these tangents of rays parallel to axis intersect in focus F.

Which gives the axis itself, then the directrix through H symetric of F from the tangent in A, or just AH=AF.

A construction by Newton :

The diagonals AC and BD intersect in O.

Circle with diameter AC intersects the perpendicular from O to AC in point K. OM = OM' = OK is copied to AC line. OM² = OM'² = OA.OC

In the same way, circle with diameter BD intersects the perpendicular from O to BD in K' and ON = OK' is copied on BD line. ON² = OB.OD

MN and M'N are the directions of axis of the two parabolas going through points ABCD.

The Pascal theorem and the point at infinity in direction MN as a 5^{th} point,
give a construction of the tangents to parabola in A and D.

Parallel to MN from C intersects AD in U.
UO intersects the parallel from B to MN in point V.
AV is the tangent in A.

The tangent in D is similarily constructed, exchanging A↔D and B↔C, DV' is the tangent in D.

The parallels to MN from A and D reflect on the tangents into rays going at focus F.

The symetric H of F from the tangent in A (or just AH = AF) is on the directrix,
which is then the perpendicular to MN from H.

The tangent at apex is then the Simson line of the focus for triangle ABC, hence the focus is on the circumcircle of ABC.

Hence the construction :

from C draw the perpendicular to axis direction.
It intersects the circumcircle in P.

The perpendicular from P to AB intersects the circumcircle at the focus F.

The symetric of F from AB gives a point K on the directrix,
the perpendicular from K to axis direction is then the directrix.

The parallel from K to axis direction gives the contact point T on AB.

Symetric points of F from BC and AC give the other contact points.

The points on the directrix define the directrix as the Steiner line of focus.
This line (that is the directrix) goes at orthocenter H of ABC (as Steiner line).

This gives another construction : the directrix is the perpendicular to axis direction
from the orthocenter H of ABC.

Focus is then the intersection point of the symetric lines of directrix from the sides (the tangents).

Hence the construction : circumcircles to ABC and ADE intersect in F.

Symetric points H and K of F from the two tangents AB and AD define the directrix.

Perpendiculars to the directrix from these points give the contact points T1 and T4, and similarily for the two other tangents.

The orthocenters of the 4 triangles are in line on the Steiner line.

This gives a direct construction of the directrix by constructing the orthocenters
H1 and H2 of triangles CDG and EBG,

then the symetric lines of directrix from the sides
AB and DE intersect at focus F.

For instance, given the

Unfortunately, this requires a 3

More details about this case.

And also : construct a parabola with given focus.